A proton enters a parallel-plate capacitor traveling to the right at a speed of
ID: 1573044 • Letter: A
Question
A proton enters a parallel-plate capacitor traveling to the right at a speed of 1.272 x 10-5 m/s, as shown in the figure. The distance between the two plates is 1.57 cm. The proton enters the capacitor halfway between the top plate and the bottom plate; that is, a distance r = 0.785 cm from each plate, as shown in the figure. The capacitor has a 2.75 10.4 N/C uniform electric field between the plates that points downward from the top plate to the bottom plate. Neglecting gravitational forces, what horizontal distance does the proton traverse before the proton hits the bottom plate? 9.53e-9 X How does force relate to the electric field? Can you solve Newton's second law for acceleration? What kinematic equation can you use to determine the final velocity? m Additional MaterialsExplanation / Answer
Force on proton,
F = qE
and F = ma (From neuton's second law)
so, acceleration a = qE / m
a = [1.6*10^(-19)* 2.75*10^(-4)] / [1.67*10^(-27)]
a = 26347.30 m/s^2
From kinematic equation,
r = ut + (1/2)at^2
where, t = time required to hit the bottom plate
u = initial speed = 0
so, t = sqrt (2r / a)
t = sqrt (2*0.785*10^(-2) / 26347.30 )
t = 7.71*10^(-4) s
Horizontal distance travelled to hit the bottom plate,
R = vx * t
R = 1.272*10^(-5) * 7.71*10^(-4)
R = 9.82*10^(-9) m